Question

A lawn mower run on a mixture of gas and oil. The ratio of ounces of gas to ounces of oil is 32 to 1, how many ounces of oil does the lawn mower need for 64 ounces of gass

Answer 1

Since the ratio should be linear, we will solve as follows:

We multiply the new ratio for gas (64 ounces) times the initial ratio of oil (1 ounce) and divide by the original ratio of gas (32 ounces), that is:

$x=\frac{64\cdot1}{32}\Rightarrow x=2$

So, we will have that we will need 2 ounces of oil.